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For which value(s) of $p$ does $f(x)=p$ have:

  • no solutions

  • exactly 1 solution

  • exactly 2 solutions

  • exactly 3 solutions

So, I have one problem which keeps me from solving this problem: how do I find the beginning point of this function? I have a graphing calculator, with the ability to calculate 'value, zero, minimum, maximum, intersect and dy/dx'. How can one find out what the beginning of this function is (with a graphing calculator or with algebra). I don't think I'll have trouble with this problem once I know the beginning point.

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which values of what? $x$ or $p$? And you have to specify if the solutions being referred to are complex or real. – user31280 Oct 25 '12 at 15:34
Is it correct now? Or for which $p\epsilon\mathbb{R}$ – JohnPhteven Oct 25 '12 at 15:36
up vote 3 down vote accepted

The function fails to be be defined when the argument in the square root takes negative values, i.e. for $x< -2.5$. Then, for $x\geq -2.5$, you want to count the solutions to $p=x^{2}\sqrt{2x+5}-6$ for each $p$. The easiest way to see how this changes is to graph it, fix a value of $p$ on the $y$-axis, extend a horizontal line through that point, and count the intersections. You'll see theres an interval $(b,\infty)$ on which it has one solution, two points where it has two solutions, and an interval $(a,b)$ on which it has three solutions.

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$ x\geq -2,5 $, so beginning punt is $(-2,5; f(-2,5))$ – JohnPhteven Oct 25 '12 at 15:31
That's correct. – user123123 Oct 25 '12 at 15:32
Thank you, problem solved! – JohnPhteven Oct 25 '12 at 15:34

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